Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<iostream>
#include<queue>
#include<vector>
using namespace std;

#define MAX_N 500000
#define INF 999999999999999

vector<long long>x[MAX_N];
long long d[MAX_N];
long long n,m,a,b,s,t,q;
queue<long long>Q;
long long sum1[MAX_N];
long long sum2[MAX_N];
long long sum;

void bfs(int P){
	for(int i=0;i<MAX_N;i++){
		d[i]=INF;
	}
	d[P]=0;
	Q.push(P);
	while(!Q.empty()){
		int w=Q.front();Q.pop();
		for(int i=0;i<x[w].size();i++){
			if(d[x[w][i]]==INF){
				d[x[w][i]]=d[w]+1;
				Q.push(x[w][i]);
			}
		}
	}
}

int main(){
	cin>>n>>m>>s>>t;
	for(int i=0;i<m;i++){
		cin>>a>>b;
		x[a].push_back(b);
		x[b].push_back(a);
	}
	bfs(s);
	q=d[t];
	for(int i=1;i<=n;i++){
		if(d[i]<INF){
			sum1[d[i]]++;
		}
	}
	bfs(t);
	for(int i=1;i<=n;i++){
		if(d[i]<INF){
			sum2[d[i]]++;
		}
	}
	for(int i=0;i<q-1;i++){
		sum+=sum1[i]*sum2[q-i-2];
	}
	cout<<sum<<endl;
	return 0;
}